\newproblem{lay:1_1_11}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.1.11}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Solve the equation system
	\begin{center}
		$\begin{array}{rcr}
			x_2+5x_3&=&-4\\
			x_1+4x_2+3x_3&=&-2 \\
			2x_1+7x_2+1x_3&=&-2
		\end{array}$
	\end{center}
}{
   % Solution
	Let us construct the augmented system matrix
	\begin{center}
		$\left(\begin{array}{rrr|r}
		   0 & 1 & 5 & -4 \\
			 1 & 4 & 3 & -2 \\
			 2 & 7 & 1 & -2
		\end{array}\right)$
	\end{center}
	Now, we apply row operations to solve it
	\begin{center}
		\begin{tabular}{cc}
			 $\mathbf{r}_2\leftrightarrow \mathbf{r}_1$ &
			 $\left(\begin{array}{rrr|r}
				 1 & 4 & 3 & -2 \\
				 0 & 1 & 5 & -4 \\
				 2 & 7 & 1 & -2
			 \end{array}\right)$ \\
			 $\mathbf{r}_3\leftarrow \mathbf{r}_3-2\mathbf{r}_1$ &
			 $\left(\begin{array}{rrr|r}
				 1 & 4 & 3 & -2 \\
				 0 & 1 & 5 & -4 \\
				 0 & -1 & -5 & 2
			 \end{array}\right)$ \\
			 $\mathbf{r}_3\leftarrow \mathbf{r}_3+\mathbf{r}_2$ &
			 $\left(\begin{array}{rrr|r}
				 1 & 4 & 3 & -2 \\
				 0 & 1 & 5 & -4 \\
				 0 & 0 & 0 & -2
			 \end{array}\right)$ \\
		\end{tabular}
	\end{center}
  Last row represents the equation $0=-2$ which is non-sense and, therefore, there is no solution of the system. The equation system is incompatible.
}
\useproblem{lay:1_1_11}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
